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Sagot :
To determine the standard deviation of the sample data summarized in the frequency distribution table, follow these steps:
1. Recognize the Class Intervals and Frequencies:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Interval} & 30-36 & 37-43 & 44-50 & 51-57 & 58-64 & 65-71 \\ \hline \text{Frequency} & 2 & 24 & 44 & 15 & 6 & 4 \\ \hline \end{array} \][/tex]
2. Calculate the Class Midpoints:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2} \][/tex]
- For [tex]\(30-36\)[/tex]: [tex]\(\frac{30 + 36}{2} = 33\)[/tex]
- For [tex]\(37-43\)[/tex]: [tex]\(\frac{37 + 43}{2} = 40\)[/tex]
- For [tex]\(44-50\)[/tex]: [tex]\(\frac{44 + 50}{2} = 47\)[/tex]
- For [tex]\(51-57\)[/tex]: [tex]\(\frac{51 + 57}{2} = 54\)[/tex]
- For [tex]\(58-64\)[/tex]: [tex]\(\frac{58 + 64}{2} = 61\)[/tex]
- For [tex]\(65-71\)[/tex]: [tex]\(\frac{65 + 71}{2} = 68\)[/tex]
3. Calculate the Mean of the Distribution:
[tex]\[ \text{Mean} (\bar{x}) = \frac{\sum f \cdot x}{\sum f} \][/tex]
Where [tex]\(f\)[/tex] is the frequency, and [tex]\(x\)[/tex] is the class midpoint.
[tex]\[ \begin{align*} \sum f \cdot x &= 2 \cdot 33 + 24 \cdot 40 + 44 \cdot 47 + 15 \cdot 54 + 6 \cdot 61 + 4 \cdot 68 \\ &= 66 + 960 + 2068 + 810 + 366 + 272 \\ &= 4542 \end{align*} \][/tex]
[tex]\[ \sum f = 2 + 24 + 44 + 15 + 6 + 4 = 95 \][/tex]
[tex]\[ \text{Mean} (\bar{x}) = \frac{4542}{95} \approx 47.81 \][/tex]
4. Calculate the Variance:
[tex]\[ s^2 = \frac{\sum f(x - \bar{x})^2}{\sum f - 1} \][/tex]
Where [tex]\((x - \bar{x})\)[/tex] is the deviation of the midpoint from the mean.
[tex]\[ \begin{align*} \sum f(x - \bar{x})^2 &= 2(33 - 47.81)^2 + 24(40 - 47.81)^2 + 44(47 - 47.81)^2 + 15(54 - 47.81)^2 + 6(61 - 47.81)^2 + 4(68 - 47.81)^2 \\ &= 4693.632 \end{align*} \][/tex]
[tex]\[ s^2 = \frac{4693.632}{94} \approx 55.11 \][/tex]
5. Calculate the Standard Deviation:
[tex]\[ s = \sqrt{s^2} = \sqrt{55.11} \approx 7.4 \][/tex]
Therefore, the standard deviation [tex]\( s \)[/tex] of the sample data is approximately [tex]\( 7.4 \)[/tex] when rounded to one decimal place.
1. Recognize the Class Intervals and Frequencies:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Interval} & 30-36 & 37-43 & 44-50 & 51-57 & 58-64 & 65-71 \\ \hline \text{Frequency} & 2 & 24 & 44 & 15 & 6 & 4 \\ \hline \end{array} \][/tex]
2. Calculate the Class Midpoints:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2} \][/tex]
- For [tex]\(30-36\)[/tex]: [tex]\(\frac{30 + 36}{2} = 33\)[/tex]
- For [tex]\(37-43\)[/tex]: [tex]\(\frac{37 + 43}{2} = 40\)[/tex]
- For [tex]\(44-50\)[/tex]: [tex]\(\frac{44 + 50}{2} = 47\)[/tex]
- For [tex]\(51-57\)[/tex]: [tex]\(\frac{51 + 57}{2} = 54\)[/tex]
- For [tex]\(58-64\)[/tex]: [tex]\(\frac{58 + 64}{2} = 61\)[/tex]
- For [tex]\(65-71\)[/tex]: [tex]\(\frac{65 + 71}{2} = 68\)[/tex]
3. Calculate the Mean of the Distribution:
[tex]\[ \text{Mean} (\bar{x}) = \frac{\sum f \cdot x}{\sum f} \][/tex]
Where [tex]\(f\)[/tex] is the frequency, and [tex]\(x\)[/tex] is the class midpoint.
[tex]\[ \begin{align*} \sum f \cdot x &= 2 \cdot 33 + 24 \cdot 40 + 44 \cdot 47 + 15 \cdot 54 + 6 \cdot 61 + 4 \cdot 68 \\ &= 66 + 960 + 2068 + 810 + 366 + 272 \\ &= 4542 \end{align*} \][/tex]
[tex]\[ \sum f = 2 + 24 + 44 + 15 + 6 + 4 = 95 \][/tex]
[tex]\[ \text{Mean} (\bar{x}) = \frac{4542}{95} \approx 47.81 \][/tex]
4. Calculate the Variance:
[tex]\[ s^2 = \frac{\sum f(x - \bar{x})^2}{\sum f - 1} \][/tex]
Where [tex]\((x - \bar{x})\)[/tex] is the deviation of the midpoint from the mean.
[tex]\[ \begin{align*} \sum f(x - \bar{x})^2 &= 2(33 - 47.81)^2 + 24(40 - 47.81)^2 + 44(47 - 47.81)^2 + 15(54 - 47.81)^2 + 6(61 - 47.81)^2 + 4(68 - 47.81)^2 \\ &= 4693.632 \end{align*} \][/tex]
[tex]\[ s^2 = \frac{4693.632}{94} \approx 55.11 \][/tex]
5. Calculate the Standard Deviation:
[tex]\[ s = \sqrt{s^2} = \sqrt{55.11} \approx 7.4 \][/tex]
Therefore, the standard deviation [tex]\( s \)[/tex] of the sample data is approximately [tex]\( 7.4 \)[/tex] when rounded to one decimal place.
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